Question

在mysql中更新时如何使用连接？到目前为止，我有这个......

$q = $dbc -> prepare("UPDATE items i JOIN accounts a SET i.shrapnel = i.shrapnel-1000, a.gender = a.? WHERE id = ? LIMIT 1");
$q -> execute(array($gender, $user['id']));

我对此非常陌生，我做错了什么？

Answer 1

提供了这些工作

如果id字段同时位于items和accounts表
id与两个表中的数据类型相同

你可以尝试

$q = $dbc -> prepare("UPDATE items i JOIN accounts a USING (id) SET i.shrapnel = i.shrapnel-1000, a.gender = a.? WHERE i.id = ? LIMIT 1");
$q -> execute(array($gender, $user['id']));

也

$q = $dbc -> prepare("UPDATE items i JOIN accounts a ON i.id = a.id SET i.shrapnel = i.shrapnel-1000, a.gender = a.? WHERE i.id = ? LIMIT 1");
$q -> execute(array($gender, $user['id']));

也

$q = $dbc -> prepare("UPDATE items i NATURAL JOIN accounts a SET i.shrapnel = i.shrapnel-1000, a.gender = a.? WHERE i.id = ? LIMIT 1");
$q -> execute(array($gender, $user['id']));

Answer 2

id需要一个表标识符。我也会用“AS”。

$q = $dbc -> prepare("UPDATE items AS i JOIN accounts AS a SET i.shrapnel = i.shrapnel-1000, a.gender = a.? WHERE a.id = ? LIMIT 1");
$q -> execute(array($gender, $user['id']));

Answer 3

您可能需要加入开启某些内容吗？：

$q = $dbc -> prepare("
UPDATE 
  items i JOIN accounts a ON (i.commonField = a.commonField) 
SET 
  i.shrapnel = i.shrapnel-1000 , 
  a.gender = a.? 
WHERE 
  id = ? LIMIT 1");
$q -> execute(array($gender, $user['id']));

检查：http://dev.mysql.com/doc/refman/5.0/en/join.html

JOIN UPDATE mysql，无法让它工作

3 个答案: